The outer electronic configuration of two members of the lanthanoid series are as follows:
`4f^(1)5d^(1)6s^(2)` and `4f^(7) 5d^(0) 6s^(2)`
What are their atomic number ? Predict the oxidation states exhibited by elements in their compounds.

Answer

Complete E.C. of 1st lanthanoid=`[Xe]^(54)4f^(1)5d^(1)6s^(2)`. Atomic No. =58(Ce)
Complete E.C. of 2nd lanthanoid=`[Xe]^(54)4f^(7)5d^(0)6s^(2)`. Atomic No. =63(Eu)
Oxidation state of 1st lanthanoid=`+2(4f^(2)+3 )4f^(1)+4(4f^(@))`
oxidation state of 2nd lanthanoid=`+2(4f^(7))` and `+3(4f^(6)`.